What is the integral of 1/(x(2x+3)) ?

The integral of 1/(x(2x+3)) is -1/3 (ln| 4/3 x+2|-ln(4/3 |x|))+C

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integral of 1/(x(2x+3))

\int \frac{1}{x ( 2 x + 3 )} d x

- \frac{1}{3} (ln \frac{4}{3} x + 2 - ln (\frac{4}{3} ∣ x ∣)) + C

Solution steps

\int \frac{1}{x ( 2 x + 3 )} d x

Complete the square

x (2 x + 3) : 2 (x + \frac{3}{4})^{2} - \frac{9}{8}

= \int \frac{1}{2 ( x + \frac{3}{4} ) ^{2} - \frac{9}{8}} d x

Apply u-substitution

: \int \frac{8}{16 u ^{2} - 9} d u

= \int \frac{8}{16 u ^{2} - 9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{16 u ^{2} - 9} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{12 ( v ^{2} - 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{12} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 8 \cdot \frac{1}{12} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{12} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 8 \cdot \frac{1}{12} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 8 \cdot \frac{1}{12} (- (\frac{ln \frac{4}{3} ( x + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{3} ( x + \frac{3}{4} ) - 1}{2}))

Simplify

8 \cdot \frac{1}{12} (- (\frac{ln \frac{4}{3} ( x + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{3} ( x + \frac{3}{4} ) - 1}{2})) : - \frac{1}{3} (ln \frac{4}{3} x + 2 - ln (\frac{4}{3} ∣ x ∣))

= - \frac{1}{3} (ln \frac{4}{3} x + 2 - ln (\frac{4}{3} ∣ x ∣))

Add a constant to the solution

= - \frac{1}{3} (ln \frac{4}{3} x + 2 - ln (\frac{4}{3} ∣ x ∣)) + C

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What is the integral of 1/(x(2x+3)) ?
The integral of 1/(x(2x+3)) is -1/3 (ln| 4/3 x+2|-ln(4/3 |x|))+C